According to this discussion https://mathoverflow.net/questions/42215/does-constructing-non-measurable-sets-require-the-axiom-of-choice, in order to construct a non lebesgue measurable set this requires the axiom of choice (via results of Solovay and Shelah). This is my understanding.
However, a friend of mine just showed me a proof of a non lebesgue measurable set assuming only CH, not AC. The same proof my friend showed me is discussed here: Lebesgue nonmeasurable sets
Also here is a comment of someone claiming to be able to make a non lebesgue measurable set using something weaker than AC: https://math.stackexchange.com/q/3791000
I know GCH implies AC and hence non lebesgue measurable sets. However is CH alone enough to give you non lebesgue measurable sets?
Without the Axiom of Choice the question becomes what do you mean by $\sf CH$. In Solovay's model the perfect set property holds, namely every uncountable set of reals contains a copy of a Cantor set, and in particular, any uncountable set of reals has size continuum. So in that sense, the Continuum Hypothesis holds.
On the other hand, requiring that $|\Bbb R|=\aleph_1$ means that the real numbers can be well-ordered, pretty much by definition. This provides us with all the choice we need to repeat any of the usual constructions of non measurable sets. From Vitali sets, to ultrafilters on $\Bbb N$, to Hahn–Banach and Banach–Tarski.
What Shelah showed is that in fact just requiring that $\aleph_1\leq|\Bbb R|$ is enough to construct a non measurable set. Although it might not be enough to get some of the aforementioned "traditional" examples.